Happy new year 2019 digit problem

apple2357

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The number 2019, its double 4038, and its triple 6057 contain all 10 digits. I am interested in the next number with this property?
I have found one but am not sure if there is one before it. My approach has been trial and error but feel there ought to be some kind of strategy. I am guessing the second digit 0?

Any thoughts?
 

Jomo

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The number 2019, its double 4038, and its triple 6057 contain all 10 digits. I am interested in the next number with this property?
I have found one but am not sure if there is one before it. My approach has been trial and error but feel there ought to be some kind of strategy. I am guessing the second digit 0?

Any thoughts?
Which number did you find? How do you suppose to show that there is not one before it?
 

Denis

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Belated Xmas gift (18 cases):

1692 3384 5076
1809 3618 5427
1902 3804 5706
1908 3816 5724
1920 3840 5760
2019 4038 6057
2079 4158 6237
2169 4338 6507
2190 4380 6570
2673 5346 8019
2697 5394 8091
2703 5406 8109
2730 5460 8190
2967 5934 8901
3027 6054 9081
3078 6156 9234
3267 6534 9801
3270 6540 9810

Brute strength. I see no way to nicely "calculate".
 

Jomo

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Belated Xmas gift (18 cases):

1692 3384 5076
1809 3618 5427
1902 3804 5706
1908 3816 5724
1920 3840 5760
2019 4038 6057
2079 4158 6237
2169 4338 6507
2190 4380 6570
2673 5346 8019
2697 5394 8091
2703 5406 8109
2730 5460 8190
2967 5934 8901
3027 6054 9081
3078 6156 9234
3267 6534 9801
3270 6540 9810

Brute strength. I see no way to nicely "calculate".
The sum of the digits of all the numbers in column 3 add to 18. Why is that?
 

Denis

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The sum of the digits of all the numbers in column 3 add to 18. Why is that?
No idea...good stuff for your "whiteboard" :p

Seems to be for that "case" only; tried a few others,
like 2789: 2789*3 = 8367: sum digits = 24
 

Jomo

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No idea...good stuff for your "whiteboard" :p

Seems to be for that "case" only; tried a few others,
like 2789: 2789*3 = 8367: sum digits = 24
But this is not the special number that fits our case.
 

Denis

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But this is not the special number that fits our case.
That's what I meant with "Seems to be for that "case" only".

I was unclearly saying that the "18" result is only when all 10 digits appear.
 

apple2357

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Which number did you find? How do you suppose to show that there is not one before it?

The one i got was 3027! But it was mostly guess work!
 

Denis

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Any "reason" you posted this?
 

apple2357

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Any "reason" you posted this?
Someone sent it to me. And i couldn't see an interesting way to approach it so wondered if i was missing something!
 

Jomo

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That's what I meant with "Seems to be for that "case" only".

I was unclearly saying that the "18" result is only when all 10 digits appear.
Denis, can you please do me a favor and run your program for more numbers of this type?
 

Subhotosh Khan

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Denis, can you please do me a favor and run your program for more numbers of this type?
If you play LOTTO with those numbers - he can demand a cut of the winnings!!
 

Jomo

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Denis

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Denis, can you please do me a favor and run your program for more numbers of this type?
There's no more.

Didn't write a program: calculated those in my head during my morning walk :lol:
 
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Jomo

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1234567890 2469135780 3703703670-----sum of digits of the last number is 36, which is 2*18. Strange

9012345678 ..... ..... sum of digits of the last number is 36, which is 2*18. Strange

If this pattern of multiples of 18 will continue, then I will like to try to prove this with minimal help. So if someone sees the prove, can you please give me some leading hints.
 
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Jomo

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There's no more.

Didn't write a program: calculated those in my head during my morning walk :lol:
What if I say please?? Please run your program. And I was joking about you being a demanding person.
 

Denis

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What do you mean Steve?
There are no more cases of 3 4digit numbers
containing all the 10 digits (at least once each),
where (1st number = a) the 3 numbers are a, 2a and 3a.
 

Jomo

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What do you mean Steve?
There are no more cases of 3 4digit numbers
containing all the 10 digits (at least once each),
where (1st number = a) the 3 numbers are a, 2a and 3a.
I guess what you are saying is true but there are 5, 6, 7, .... digit numbers which you did not find. This really seems like an interesting problem even if does not go on for ever and I'd like to look into this. After all, this is what real mathematicians do (not that I am one).
 

Denis

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I guess what you are saying is true but there are 5, 6, 7, .... digit numbers which you did not find.
BUT I was NOT looking for those: I was looking for 3 numbers that are
each 4 digits, NOTHING else. Which meant the lower was in range
1000 to 3333, meaning high case of 3333,6666,9999.
Sorry 'bout dat, chief...as Maxwell Smart would say...
 

Jomo

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BUT I was NOT looking for those: I was looking for 3 numbers that are
each 4 digits, NOTHING else. Which meant the lower was in range
1000 to 3333, meaning high case of 3333,6666,9999.
Sorry 'bout dat, chief...as Maxwell Smart would say...
Look Jackson, I am asking you for a favor. Can you modify your morning walk a bit, please?
 
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